jmc

Let $m$ and $n$ be the degrees of $f(x)$ and $g(x),$ respectively. Then the degree of $f(g(x))$ is $mn.$ The degree of $f(x)g(x)$ is $m+n,$ so $$mn=m+n.$$Applying Simon's Favorite Factoring Trick, we get $(m-1)(n-1)=1,$ so $m=n=2.$

Let $f(x)=a{x}^2+bx+c$ and $g(x)=d{x}^2+ex+f.$ Then $$a(d{x}^2+ex+f{)}^2+b(d{x}^2+ex+f)+c=(a{x}^2+bx+c)(d{x}^2+ex+f).$$Expanding, we get $$\begin{array}{rl}& a{d}^2{x}^4+2ade{x}^3+(2adf+a{e}^2+bd)x^2+(2aef+be)x+a{f}^2+bf+c\\ & =ad{x}^4+(ae+bd)x^3+(af+be+cd)x^2+(bf+ce)x+cf.\end{array}$$Matching coefficients, we get $$\begin{array}{rl}a{d}^2& =ad,\\ 2ade& =ae+bd,\\ 2adf+a{e}^2+bd& =af+be+cd,\\ 2aef+be& =bf+ce,\\ a{f}^2+bf+c& =cf.\end{array}$$Since $a$ and $d$ are nonzero, the equation $a{d}^2=ad$ tells us $d=1.$ Thus, the system becomes $$\begin{array}{rl}2ae& =ae+b,\\ 2af+a{e}^2+b& =af+be+c,\\ 2aef+be& =bf+ce,\\ a{f}^2+bf+c& =cf.\end{array}$$Then $b=ae.$ Substituting, the system becomes $$\begin{array}{rl}2af+a{e}^2+ae& =af+a{e}^2+c,\\ 2aef+a{e}^2& =aef+ce,\\ a{f}^2+aef+c& =cf.\end{array}$$Then $af+ae=c,$ so $a{f}^2+aef=cf$. Hence, $c=0,$ which means $ae+af=0.$ Since $a$ is nonzero, $e+f=0.$

Now, from $g(2)=37,$ $4+2e+f=37.$ Hence, $e=33$ and $f=-33.$ Therefore, $g(x)=\boxed{x^2+33x-33}.$